Show that sum of first n even natural numbers is equal to (n+1/n)times the sum of first n odd natural numbers.

Se= 2+ 4+…+ 2n =2(1+2+3+…+n) = 2* n(n+1)/2 = n(n+1)

The sum of the first N odd natural numbers

So= 1+3 +…+ (2n-1) [ arthemic series with common difference 2 ]

= n(1+2n-1)/2 = n^2.

We see that (1+ 1/n)So = (1+ 1/n)n^2 = n^2 + n = n(n+1) =Se

Hence, proved.